Difference between revisions of "1983 AHSME Problems/Problem 25"
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Therefore, we have | Therefore, we have | ||
<cmath>60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}</cmath> | <cmath>60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}</cmath> | ||
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+ | == Solution 2 == | ||
+ | We have <math>60^a = 3</math> and <math>60^b = 5</math>. We can say that <math>a = log_{60} 3</math> and <math>b = log_{60} 5</math>. | ||
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+ | <cmath>12^{(1-a-b)/[2(1-b)]} = 12^{(1-(a+b))/[2(1-b)]}</cmath> | ||
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+ | We can evaluate (a+b) by the Addition Identity for Logarithms, <math>(a+b) = log_{60} 15</math>. Also, <math>1 = log_{60} 60</math>. | ||
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+ | <cmath> (1-(a+b) = log_{60} 60 - log_{60} 15 = log_{60} 4 </cmath> | ||
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+ | Now we have to evaluate [2(1-b)], the denominator of the fractional exponent. Once again, we can say <math>1 = log_{60} 60</math> | ||
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+ | <cmath> 2(1-b) = 2(log_{60} 12)</cmath> | ||
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+ | <cmath>12^{(log_{60} 4)/[2(log_{60} 12]} = 12^{\frac{1}{2} \cdot log_{12} 4} = 4^{1/2} = 2</cmath> | ||
==See Also== | ==See Also== |
Revision as of 11:12, 14 January 2022
Contents
Problem 25
If and , then is
Solution
We have that . We can substitute our value for 5, to get Hence Since , we have Therefore, we have
Solution 2
We have and . We can say that and .
We can evaluate (a+b) by the Addition Identity for Logarithms, . Also, .
Now we have to evaluate [2(1-b)], the denominator of the fractional exponent. Once again, we can say
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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